I am trying solve the following problem: Fix $0 < \gamma < 1$ and suppose $f$ is holomorphic on the unit disc satisfying $$|f(z)-f(w)| \le C_1|z-w|^\gamma$$ for some constant $C_1$. Show that there exists a constant $C_2 > 0$ s.t $$|f'(z)| \le C_2(1 - |z|)^{\gamma - 1}$$.
My attempt: Since we want to bound the derivative of $f$, I thought it makes sense to use the Cauchy's Integral Formula, namely we have: $$|f'(z)| = \frac{1}{2\pi}|\int_{C}\frac{f(w)}{(w-z)^2}dw|,$$ for any circle $C$ around the origin of radius $r < 1$. Using the assumption, we have
\begin{array}{lcl} |f'(z)| & \le & \frac{1}{2\pi}\int_{C}|\frac{f(w)}{(w-z)^2}|dw \\ & \le & \frac{1}{2\pi}\int_{C}|\frac{f(w)C_1^{\frac{2}{\gamma}}}{(f(w)-f(z))^{\frac{2}{\gamma}}}| \\ \end{array}
However, this does not look promising, we need to get an exponent $\gamma - 1$. Can someone help me, please?
Thanks, in advance.